# Find the area of a triangle ABC, where AB = 6.8m, BC = 9.5m and AC = 10.2m.

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The cosine of a triangle is given by;

`cosA = (b^2+c^2-a^2)/(2bc)`

A is the angle between AB and AC.

a = Opposite leg to the angle A.(leg BC) = 9.5cm

b = Opposite leg to the angle B.(leg AC) = 10.2cm

c = Opposite leg to the angle C.(leg AB) = 6.8cm

`cosA = (10.2^2+6.8^2-9.5^2)/(2*10.2*6.8) = 0.433`

We know that ` sin^2A+cos^2A = 1`

Therefore;

`sinA = sqrt(1-cos^2A) = sqrt(1-0.433^2) = 0.813`

If we assume that the perpendicular drawn from point B will meet AC at point D;

`AD = ABsinA = 6.8*0.813 = 5.528`

Area of triangle

= 1/2(base length*perpendicular height)

`= 1/2*AC*AD`

`= 1/2*10.2*5.528`

`= 28.193cm^2`

** So the area of the triangle is **`28.193cm^2`

To find the area of the triangle, given the length of the three sides, we may use the Heron's Formula.

`A = sqrt(s(s-a)(s-b)(s-c))`

where A - Area of triangle

a, b, c - length of the sides of the triangle

s - semi-perimeter, `s = (a+b+c)/2`

So, let a =AB=6.8m, b = BC=9.5m and c=AC=10.2m.

Then, solve for s.

`s = (a+b+c)/2 = (6.8 + 9.5+10.2)/2 = 26.5/2= 13.25`

And, substitute values of , b, c and s to the formula of area of triangle.

`A=sqrt(13.25(13.25-6.8)(13.25-9.5)(13.25-10.2))`

`A=sqrt(13.25(6.45)(3.75)(3.05))`

`A=sqrt977.4773437`

`A=31.26`

**Hence, the area of triangle is `31.26m^2` .**